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college physics

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A certain 9-V battery measures 9.0 V with no load connected to it, but only 8.5 V when connected to a radio which draws 150 mA of current.

What is the internal resistance of the battery?

  • college physics - ,

    .150*internalresistance=9-8.5

  • college physics - ,

    I see you're basically finding the voltage drop that occurs as a result of that internal resistance.

    Now they ask, in the second part .... What would be the terminal voltage for a 50-W load?

  • college physics - ,

    Oops. question should say 50 ohm load NOT 50-W load. Sorry.


    Terminal Voltage as I understand it would be the Vt = Vo - I(Ri)

    Where Vt is terminal voltage, Vo is battery voltage, I is current and Ri is internal resistance. Is that correct? How would they do it?

  • college physics - ,

    current = 9.0/(internal resisitance + 50)
    then
    terminal voltage= 9.0-current*Ri

  • college physics - ,

    thank you very much

  • college physics - ,

    sorry here's another question ... they also ask what is the maximum amount of power you can get out of this battery?

    You have P = IV. Max voltage is 9V, and max current is 2.7A. Then in that case that would be 9 * 2.7 = 24.3. Did I do that right?

  • college physics - ,

    No, not at all. Max power is delived when load resistance is equal to internal resistance.


    Proof: Load power= Rl*Il^2=Rl(9/(Ri+Rl)^2

    take the derivative:

    9/(Ri+Rl)^2 - 18Rl/(Ri+Rl)^3
    set that equal to zero, and solve for Rl

    Rl=1/2 (Ri+Rl)
    Rl=Ri for max power.

  • college physics - ,

    Yes, you short circuited the terminals. In all probability the battery will not last long but it sure will get hot.

  • college physics - ,

    However as far as useful power out of the battery instead of generating heat inside you had best look for where the external voltage times the current ins maximum.

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